Starch and
cellulose have the same substance but different structures. They are both
polysaccharides. The basic unit of a polysaccharide is the glucose. Glucose,
which contains carbon, hydrogen, and oxygen, have two forms. The alpha-glucose
with an alcohol group attached to carbon 1 is down and the beta-glucose with
the alcohol group attached to carbon 1 is up. Starch is the alpha-glucose while
cellulose is the beta-glucose. Starches are linked into a straight chain
whereas the cellulose are connected like a pile of stack paper. When the human
body eats starch, it can digest the starch but not the cellulose because it has
no enzyme that can break it down.
Answer:
Explanation:
For this case we can use the following formula for the angular velocity:
where represent the final angular velocity , the initial angular velocity , t the time and the angular acceleration.
And for the linear acceleration we have this formula:
Where a represent the linear acceleration and the angular acceleration.
For this case the linear acceeleration is given
And the radius of the yo-yo is also given
So then we can use the following formula:
If we replace the values we got:
And solving for we got:
"The movement of water into a nutrient-rich region of the phloem decreases the pressure in that region" is the statement that is not true according <span>to the pressure-flow hypothesis. The correct option among all the options that are given in the question is the fourth option or the last option. I hope it helps you.</span>
Answer:
Maximum 13 m overhang
Explanation:
Since the brick is uniform. Its mass must be equally distributed along its length. So the center of mass is in geometric center, which is 26 / 2 = 13 m from each end of the brick.
For the brick to not tip over, its center of mass must NOT be overhung over the edge. It stays just on the edge. So the maximum overhang is 13 m.
Answer:
Explanation:
Given the following :
Temperature (Th) of hot reservoir = 30°C (30 +273) = 303K
Temperature (Tc) of cold reservoir = 0°C (273K)
Quantity (Q) of heat transferred from hot reservoir = 400 J
Total change in entropy (ΔStotal) :
ΔStotal = ΔShot + ΔScold
ΔS = Q/T
Hot reservoir is losing 400J of heat ;
Q = - 400 J
ΔShot = Q/Th
ΔShot = - 400/303
ΔShot = - 1.32 J/K
ΔScold = Q/T
ΔScold = 400 / 273
ΔScold = 1.47 J/K
ΔStotal = ΔShot + ΔScold
ΔStotal = - 1.32 + 1.47
ΔS total = 0.15 J/K